lab-2l+（a-1）的平方=0
ab = 2
a = 1
b = 2
1ab+1（a+1）（b+1）+1(a+2)(b+2)+...+1(a+2008)(b+2008)
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + …… + 1/2009 - 1/2010
= 1-1/2010
=2009/2010
已知lab-2l+（a-1）的平方=0,求:1\ab+I\（a+1）（b+1）+1\(a+2)(b+2)+...+1\(
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